PLEASE SHOW THE WORKING
4 Answers
2. Draw a rough graph of y = log_{2}(x+5) and y=6-x.
You'll see that they intersect once. Thus the given eq. has one solution.
3. log_{1/âˆš2}(sinx) > 0
â†’log_{2(-0.5)} (sinx)>0
â†’(-1/2)log_{2}sinx>0
â†’log_{2}sinx<0
â†’sinx<1
Simultaneously sinx>0 (domain)
only one value of x satisfies the given relation in [0,pi/2] which is pi/4
Thus in [0,4pi] there will be 4 solutions.
2 in [0,Ï€]
0 :[Ï€,2Ï€] ; 2 :[2Ï€,3Ï€] ; 0:[3Ï€,4Ï€]
4. Proceeding similarly as in the previous sum, you will see that the ineq. reduces to sin(x+Î¸)â‰¥1 where cosÎ¸=1/3
Therefore, x+Î¸=Ï€/2
or, x = asin(1/3)
In [-2Ï€,2Ï€] there will again be 4 values of x.
Answer should be D.
Q. 1:- (0.2)^{25}=2^{25}x10^{-25}. Now, log_{10}2^{25}â‰ˆ7.5 which means that there are 8 digits in 2^{25}. In 10^{-25}, after the decimal point, there are 24 zeroes and one 1, that is 25 digits. So, in (.2)^{25}, the last 8 digits will be of 2^{25} and the 17 before that will all be zero. (B).
Q. 5:- x^{2}+y^{2}=12xy or, (x+2y)^{2}=16xy. Now, log_{2}(x+2y)^{2}=log_{2}(16xy)=4+log_{2}(xy). For the given range of x and y, log_{2}(x+2y)^{2} lies between 4 and 8. So, log_{2}(x+2y) lies between 2 and 4. â†’(C).
It is apparent that for the logarithm to be 3, xy=4 and the ordered pairs are (4, 1), (2, 2), (1, 4). So (D) is not correct.